#Evaluating Definite Integrals

#Table of Contents

Introduction
Fundamental Theorem of Calculus
Steps to Evaluate Definite Integrals
Key Concepts
Worked Examples
Practice Questions
Glossary
Summary and Key Takeaways

#Introduction

Evaluating a definite integral involves finding the numerical value of the integral over a specified interval. This process is crucial in calculus and has various applications in physics, engineering, and other fields.

#Fundamental Theorem of Calculus

The first fundamental theorem of calculus states that if $f$ is a continuous function on the closed interval $[a, b]$ , then:

${\int }_{a}^{b}f(x) , dx = F(b) - F(a)$

where $F$ is an antiderivative of $f$ .

#Steps to Evaluate Definite Integrals

Find the Antiderivative: Determine the indefinite integral of the function $f(x)$ , denoted as $F(x)$ . $\int f(x) , dx = F(x) + C$
Apply the Limits: Use the evaluated antiderivative to compute the difference between its values at the upper and lower limits. ${\int }_{a}^{b}f(x) , dx = {\left[F(x)\right]}_{a}^{b} = F(b) - F(a)$
Ignore the Constant of Integration: When evaluating definite integrals, the constant of integration cancels out. ${\left[F(x) + C\right]}_{a}^{b} = (F(b) + C) - (F(a) + C) = F(b) - F(a)$

### Example 1: Basic Definite Integral Evaluate the definite integral

{\int }\_{0}^{1} x^2 \, dx

.

Solution:

Find the antiderivative: $\int x^2 , dx = \frac{1}{3}x^3 + C$
Apply the limits: ${\int }_{0}^{1} x^2 , dx = \left[\frac{1}{3}x^3\right]_{0}^{1} = \frac{1}{3}(1^3) - \frac{1}{3}(0^3) = \frac{1}{3}$

#Key Concepts

Key Concept

#Key Concept: Sign of the Integral

If $f(x) > 0$ on the interval $[a, b]$ , then ${\int }_{a}^{b} f(x) , dx > 0$ .
If $f(x) < 0$ on the interval $[a, b]$ , then ${\int }_{a}^{b} f(x) , dx < 0$ .
If $f(x)$ changes sign on $[a, b]$ , the integral can be positive, negative, or zero.

### Note: Constants of Integration When calculating definite integrals, constants of integration do not affect the result since they cancel out during the evaluation step. ### Example 2: Trigonometric Integral Evaluate the definite integral

{\int }\_{-\frac{\pi }{3}}^{\frac{\pi }{4}} \cos(x) \, dx

.

Solution:

Find the antiderivative: $\int \cos(x) , dx = \sin(x) + C$
Apply the limits: $\begin{array}{rcl} {\int }_{-\frac{\pi }{3}}^{\frac{\pi }{4}} \cos(x) , dx & = & \left[\sin(x)\right]_{-\frac{\pi }{3}}^{\frac{\pi }{4}} \\ & = & \sin\left(\frac{\pi }{4}\right) - \sin\left(-\frac{\pi }{3}\right) \\ & = & \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{3}}{2}\right) \\ & = & \frac{\sqrt{2} + \sqrt{3}}{2} \end{array}$

#Worked Examples

### Example 3: Polynomial Integral Consider the function

f

defined by

f(x) = {\int }\_{0}^{x} t^2 (2 - t) \, dt

.

(a) Calculate $f(2)$ and $f(3)$ and confirm that $f(3) < f(2)$ .

Solution:

Expand the integrand and find the indefinite integral: $\int t^2 (2 - t) , dt = \int (2t^2 - t^3) , dt = \frac{2}{3}t^3 - \frac{1}{4}t^4 + C$
Calculate $f(2)$ : $\begin{array}{rcl} f(2) & = & {\int }_{0}^{2} t^2 (2 - t) , dt \\ & = & \left[\frac{2}{3}t^3 - \frac{1}{4}t^4\right]_{0}^{2} \\ & = & \left(\frac{2}{3}(2)^3 - \frac{1}{4}(2)^4\right) - \left(\frac{2}{3}(0)^3 - \frac{1}{4}(0)^4\right) \\ & = & \frac{16}{3} - 4 \\ & = & \frac{4}{3} \end{array}$
Calculate $f(3)$ : $\begin{array}{rcl} f(3) & = & {\int }_{0}^{3} t^2 (2 - t) , dt \\ & = & \left[\frac{2}{3}t^3 - \frac{1}{4}t^4\right]_{0}^{3} \\ & = & \left(\frac{2}{3}(3)^3 - \frac{1}{4}(3)^4\right) - \left(\frac{2}{3}(0)^3 - \frac{1}{4}(0)^4\right) \\ & = & \frac{18}{4} - \frac{81}{4} \\ & = & -\frac{63}{4} \end{array}$
Confirm $f(3) < f(2)$ : $-\frac{63}{4} < \frac{4}{3}$

### Example 4: Interpretation **(b) Explain why

f(3) < f(2)

is true by considering the properties of the integrand.**

Solution:

The integrand $t^2 (2 - t)$ is positive when $0 \leq t < 2$ because both $t^2$ and $(2 - t)$ are positive.
The integrand becomes negative when $t > 2$ because $t^2$ is positive, but $(2 - t)$ is negative.
Therefore, the integral from $0$ to $2$ contributes a positive value, while the integral from $2$ to $3$ contributes a negative value, reducing the overall integral value for $f(3)$ .

#Practice Questions

Practice Question

Question 1: Evaluate the definite integral ${\int }_{1}^{4} (2x + 3) , dx$ .

Answer:

Find the antiderivative: $\int (2x + 3) , dx = x^2 + 3x + C$
Apply the limits: ${\int }_{1}^{4} (2x + 3) , dx = \left[x^2 + 3x\right]_{1}^{4} = (16 + 12) - (1 + 3) = 24 - 4 = 20$

Practice Question

Question 2: Evaluate the definite integral ${\int }_{0}^{2} e^x , dx$ .

Answer:

Find the antiderivative: $\int e^x , dx = e^x + C$
Apply the limits: ${\int }_{0}^{2} e^x , dx = \left[e^x\right]_{0}^{2} = e^2 - 1$

#Glossary

Definite Integral: An integral with specified upper and lower limits, resulting in a numerical value.
Indefinite Integral: An integral without specified limits, representing a family of functions.
Antiderivative: A function $F(x)$ such that $F'(x) = f(x)$ .
Fundamental Theorem of Calculus: Relates the derivative of an integral to the original function.

#Summary and Key Takeaways

The definite integral ${\int }_{a}^{b} f(x) , dx$ can be evaluated by finding the antiderivative $F(x)$ and computing $F(b) - F(a)$ .
Constants of integration cancel out when evaluating definite integrals.
The sign of the integral depends on the sign of the integrand over the interval.
Practice with various functions to become proficient in evaluating definite integrals.

#Key Takeaways

Understand the Fundamental Theorem of Calculus: It is the backbone of evaluating definite integrals.
Master Finding Antiderivatives: This skill is crucial for solving both definite and indefinite integrals.
Apply Limits Correctly: Ensure you substitute the upper and lower limits accurately.
Recognize the Significance of the Integrand's Sign: It affects the result of the integral.

By following these guidelines and practicing regularly, you will be well-prepared to evaluate definite integrals accurately and efficiently.